10. A cannon elevated at an angle of 35º to the
horizontal fires a cannonball, which travels the path
shown in the diagram below. [Neglect air resistance and
assume the ball lands at the same height above the
ground from which it was launched.]
If the ball lands 7.0 × 10² meters from the cannon 10.
seconds after it was fired, what is the horizontal
component of its initial velocity?
(A) 7.0 m/s (B) 35 m/s (C) 49 m/s (D) 70. m/s
Bro, i put 49 m/s, but it's actually 70 m/s
the initial HORIZONTAl speed is 70 m/s
and the initial VERTICAL speed is 49 m/s
ignoring air resistance we know that the horizontal velocity is constant so you can use the formula d/t=v
700/10= 70 m/s
to find the initial vertical speed you can plug the 70m/s into the tangent function which is tan θ = vi(y-axis) / vi(x-axis)
tan 35° = vi(y-axis) / 70
rearrange it to solve for vi (y-axis)
vi(y-axis) = 70 * tan 35°
vi (y-axis) = 49 m/s
** keep in mind y axis is for the vertical
**and x axis is for the horizontal
its 70
70 m/s
To solve this problem, we can use the principles of projectile motion. First, we need to break down the velocity of the cannonball into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component changes due to the effect of gravity.
Let's assume the initial velocity of the ball is \(v_0\), the horizontal component is \(v_x\), and the vertical component is \(v_y\). Since there is no air resistance, the only force acting on the ball in the horizontal direction is the initial velocity, while the force acting in the vertical direction is due to gravity.
The horizontal component of the initial velocity (\(v_x\)) remains constant. We can find this by using the formula: \(v_x = v_0 \cdot \cos(\theta)\), where \(\theta\) is the angle of elevation (35º in this case).
The time of flight can be found using the horizontal distance traveled by the ball (\(d\)) and the horizontal component of the initial velocity. In this case, the horizontal distance is given as 7.0 × 10² meters, and the time of flight is given as 10 seconds.
\(d = v_x \cdot t\)
\(7.0 \times 10² = v_x \cdot 10\)
Now, we can substitute the value of \(v_x\) in terms of \(v_0\) and \(\theta\) into the equation:
\(7.0 \times 10² = v_0 \cdot \cos(\theta) \cdot 10\)
Finally, we need to solve for \(v_0\):
\(v_0 = \frac{{7.0 \times 10²}}{{10 \cdot \cos(\theta)}}\)
Now, let's substitute the value of \(\theta\) (35º) into the equation and calculate the value of \(v_0\):